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3 years ago

An employee earns $175 for 15 hours work. Find the constant of variation.

2 answers:

4 0

$\text{The constant of variation is a fancy word for rate of change}\\\\\text{You're trying to find the rate of change in this scenario, which would}\\\text{be your constant of varaition}\\\\\text{To find this, we need to know how much he makes each hour}\\\\\text{Find this by dividing 175 by 15:}\\\\175\div15= 11.666\\\\\text{Round to the nearest hundredths}\\\\11.67\\\\\boxed{\text{Constant of variation: 11.67}}$

In-s [12.5K]3 years ago

3 0

Answer: 2625

Step-by-step explanation:

175*15=

2625

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The graphs of the equations y=2x-7 and y-kx=7 are parallel when k equals

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$y=2x-7 \ and \ y-kx=7 \\ \\ y=2x-7 \ and \ y=kx+7 \\ \\ two \ lines \ are \ parallel \ if \ their \ slopes \ are \ the \ same : \\ \\ k= 2$

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3 years ago

Samantha wants to use her savings of $1,150 to buy shirts and watches for her family. The total price of the shirts she bought w

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3 years ago

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a band of 45 ewoks crash-landed in the forest last night. this sounds like a small problem, but the population will grow at the

kogti [31]

Given Information:

Starting population = P₀ = 45

rate of growth = 22%

Required Information:

Population every five years from this year to the year 2050 = ?

Answer:

$Year \: 2020 = P(0) = 45e^{0} = 45\\\\Year \: 2025 = P(5) = 45e^{0.22*5} = 135\\\\Year \: 2030 = P(10) = 45e^{0.22*10} = 406\\\\Year \: 2035 = P(15) = 45e^{0.22*15} = 1,220\\\\Year \: 2040 = P(20) = 45e^{0.22*20} = 3,665\\\\Year \: 2045 = P(25) = 45e^{0.22*25} = 11,011\\\\Year \: 2050 = P(30) = 45e^{0.22*30} = 33,079\\\\$

Step-by-step explanation:

The population growth can be modeled as an exponential function,

$P(t) = P_0e^{rt}$

Where P₀ is the starting population, r is the rate of growth of the population and t is the time in years.

We are given that starting population of 45 and growth rate of 22%

$P(t) = 45e^{0.22t}$

Assuming that the starting year is 2020,

Therefore, the starting population of ewoks was 45 in 2020 and increased to 33,079 by 2050 in a time span of 30 years.

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3 years ago

Katie earns 15$ each hour. Round to total time katie worked to the nearest hour and estimate her pay by using mulitplacation

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Answer:

How long did she work for though?

Step-by-step explanation:

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2 years ago

A giant tank in a shape of an inverted cone is filled with oil. the height of the tank is 1.5 metre and its radius is 1 metre. t

skad [1K]

The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate

of dripping of 110 cm³/s gives the following values.

1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s

2) The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s

3) The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m/s

<h3>How can the rate of change of the radius & height be found?</h3>

The given parameters are;

Height of the tank, h = 1.5 m

Radius of the tank, r = 1 m

Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s

$1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h$

From the shape of the tank, we have;

$\dfrac{h}{r} = \dfrac{1.5}{1}$

Which gives;

h = 1.5·r

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}$

$\dfrac{d}{dr} V =\dfrac{d}{dr} \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi \cdot r^2$

$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}$

$\dfrac{dV}{dt} = 0.00011$

Which gives;

$\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi \cdot r^2}}$

When r = 0.5 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.5^2} \approx 9.34 \times 10^{-5}$

The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s

2) When the height is 20 cm, we have;

h = 1.5·r

$r = \dfrac{h}{1.5}$

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}$

r = 20 cm ÷ 1.5 = $13.\overline3$ cm = $0.1\overline3$ m

Which gives;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.1 \overline{3}^2} \approx \mathbf{1.313 \times 10^{-3}}$

$\dfrac{d}{dh} V = \dfrac{d}{dh} \left(\dfrac{4}{27} \cdot \pi \cdot h^3 \right) = \dfrac{4 \cdot \pi \cdot h^2}{9}$

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }$

$\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi \cdot h^2}{9}}}$

When the height is 20 cm = 0.2 m, we have;

$\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}$

The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s

3) The volume of the slick, V = π·r²·h

Where;

h = The height of the slick = 0.1 cm = 0.001 m

Therefore;

V = 0.001·π·r²

$\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi \cdot r}}$

When the radius is 10 cm = 0.1 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi \times 0.1} \approx \mathbf{0.175}$

The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m

Learn more about the rules of differentiation here:

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3 0

3 years ago